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Pythagorean Theorem in Three Dimensions

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Pythagorean Theorem in Three Dimensions The Pythagorean theorem can be extended to any number of dimensions, but most importantly three dimensions so we can use it to find any distance between two points in space. For this activity, we could have a model of a rectangular prism which can be split along the long diagonal and opened to reveal the long diagonal, d. This could be used as a physical model which could be labeled, rather than using just drawings, which can be difficult to follow. We can start with doing some examples with numbers, and then go to the formula after some practice. Presentation: • SAY: “We have learned how to find the distance between two points if we know the perpendicular distances between them; just use the Pythagorean theorem! Let’s use it to find this distance.” • We can use a diagram or a material. • SAY: “Here is a box, or a rectangular prism. Let’s say the dimensions of the box measure 3cm , 4cm, and 12cm. How far would it be from one corner of the box, all the way to the far end?” • SAY: “How can we find this length? We will be able to find it if it is a side of a right triangle. And it is! It is the hypotenuse of a right triangle. Can you find the right triangle?” • SAY: “So here is the right triangle. It has one of its legs as the height of the prism, and that is 12cm. What is the other leg, though? It is the diagonal of the bottom of the prism. But that diagonal is the hypotenuse of a right triangle as well! It is the hypotenuse of the right triangle that has legs 3cm and 4cm.” • SAY: “So we can use the Pythagorean theorem to find the hypotenuse of this triangle, which is also the leg of the other triangle! How much is this hypotenuse? Yes, it is 5cm. Now we can use the legs of 5cm and 12cm to find the diagonal of the box.” • We can give other examples, or students can work on their own if they wish. o As students are working, we can examine what happens when the numbers are not all perfect squares. o Students may discover this on their own, but if not, we can guide them to working in this way. • SAY: “Let’s try this example. Let’s say the side lengths are 5in., 9in., and 8in. as shown.” • SAY: “First we had to find the diagonal of the surface of the base by using those dimensions. Here, we get that the diagonal is 106in. We could give a decimal approximation for this number. But let’s think for a moment; what did we need to always do with this number? Yes, we will square it because it is also the leg of the large right triangle whose hypotenuse is the diagonal of the box. But what could be easier than squaring 106 ? It will just be 106! Rather than take these extra steps of finding the square root and just squaring it again, we can just leave it in exact form. Now our legs of 2 our big triangle are 8 and 106 . So d 2 = 82 + 106 = 64 +106 =170. Then our diagonal, d, would be 170in. Now if we want, we can give an approximate value, which is ≈13.04in.” After students have worked for a while, we can give them an example where the sides just have variables instead of numbers. This will lead them to a formula that can be used directly. They can do this independently, or we can work with them as needed • SAY: “As you have been working on these, you are following the same procedure every time, just with different numbers. Let’s say those dimensions then are any numbers, a, b, and c, with the long diagonal d being what we are trying to find.” • SAY: “What must we do first? Find the diagonal of the base. How do we do that? We square the sides, add them together and take the square root. Squaring these sides, we get a2 and b2 . How can we add these together? This is actually easiest of all; it is just a2 +b2 ! Now we take the square root. And just like when we had square roots that weren’t perfect squares, we will also leave this in square root form, a2 +b2 .” • We can label this in the diagram. • SAY: “What did we do next? We squared the two legs of the large triangle. The one leg is c, and the other is the length we just found, a2 +b2 . What did we do here? Same 2 as before, use the Pythagorean theorem with these two legs. So ( a2 +b2 ) + c2 = d 2 . 2 But ( a2 +b2 ) = a2 +b2 since the square and the square root are inverses of each other. Then we have that a2 +b2 + c2 = d 2 . It is the Pythagorean theorem for three dimensions! We could use this directly to find these distances.” Follow-Ups: • We could take the formula one step further as d = a2 +b2 + c2 to find the distance directly. • Students can find lengths of the diagonals of rectangular prisms given their side measures. • Students can prove the formula for the Pythagorean theorem in three dimensions on their own. • Students can find an unknown dimension of a rectangular prism given the other dimensions and a diagonal. • Students can find the distance between three point in space given their Cartesian coordinates. o Find the distance between the points (5,−4,10) and (3,−2,2). • Students can solve story problems involving distance in space. o A 20 ft.long pole needs to go in the back of a truck bed that measures 209"×80"×75" . Will the pole be able to fit in or will it be able to hang out? By how much does it fit or not fit? • Students can look for and research Pythagorean quadruples. Notes: • This is not a necessary formula, as one can always use the Pythagorean theorem twice to find the length. However, the formula is simple as it is a generalization and therefore can be quite useful. • Whole numbers that make a2 +b2 + c2 = d 2 true are called Pythagorean quadruples and lists can easily be found online. • Students should have enough practice before deriving the formula so that the process becomes routine, and thus the formula is an expression of something they have already internalized. If students cannot fully articulate the entire process of finding the diagonal of a prism, we should not force a formula on them before there is that understanding. • When we discover the formula, it does not matter what order we label the length, width, and height. However, the way presented here leads to the formula being in alphabetical order. • It is often helpful if students work with non-Pythagorean triples before trying to derive the formula, as this work helps to solidify the processes. • The expression a2 +b2 cannot be simplified to a +b. We can show this by substituting numbers in for a and b, say 3 and 4 respectively. 32 + 42 = 9 +16 = 25 and 3+ 4 = 7. Therefore, they are not equivalent expressions. .
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